Pseudocomplementation in (normal) subgrouplattices

Tom De Medts Marius Tarnauceanu

February 22, 2008

Abstract

The goal of this paper is to study finite groups admitting a pseudo-complemented subgroup lattice (PK-groups) or a pseudocomplemen-ted normal subgroup lattice (PKN-groups). In particular, we obtaina complete classification of finite PK-groups and of finite nilpotentPKN-groups. We also study groups with a Stone normal subgrouplattice, and we classify finite groups for which every subgroup has aStone normal subgroup lattice. Finally, we obtain a complete classifi-cation of finite groups for which every subgroup is monolithic.

MSC2000: primary 06D15, 20D30; secondary 20D15, 20D20

Key words: pseudocomplement, subgroup lattice, normal subgroup lattice,monolithic group, Stone lattice

1 Introduction

It is an interesting question in group theory in how far the structure of thesubgroup lattice of a group determines the structure of the group itself. Thisquestion in its pure form is quite old [11, 2], and M. Suzuki spent his earlyresearch years on this problem [14, 15]. Since then, many characterizationsand classifications have been obtained for groups for which the subgrouplattice or normal subgroup lattice has certain lattice-theoretic properties.The possibly most famous result in this direction is Ore’s result that a group

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is locally cyclic if and only if its lattice of subgroups is distributive [9]. Werefer to Suzuki’s book [17], Schmidt’s book [12] or the more recent book [21]by the second author for more information about this theory.

In the current paper, we investigate (finite) groups which have a subgrouplattice or normal subgroup lattice which is pseudocomplemented; these arecalled PK-groups and PKN-groups, respectively. We obtain a complete clas-sification of finite PK-groups and of finite nilpotent PKN-groups, therebyvastly improving earlier results [5, 19]. We will also investigate groups withthe property that each subgroup is itself a PKN-group, and we will givesome results about groups for which the normal subgroup lattice is a so-called Stone lattice. In particular, we will classify finite groups for whichthe normal subgroup lattice of every subgroup is a Stone lattice, and as acorollary, we obtain a complete classification of finite groups for which everysubgroup is monolithic.

Let (L,∧,∨) be a bounded lattice with top 1 and bottom 0, and let a ∈ L.An element b ∈ L is called a complement of a if a∨ b = 1 and a∧ b = 0, andthe lattice L is called complemented if every element of L has a complement.An element a∗ ∈ L is called a pseudocomplement of a if the following twoconditions are satisfied:

(i) a ∧ a∗ = 0 ;

(ii) a ∧ x = 0 (x ∈ L) implies x ≤ a∗ .

Any element of L can have at most one pseudocomplement. We say that L isa pseudocomplemented lattice if every element of L has a pseudocomplement.Note that the terminology is slightly misleading, since a complement is notnecessarily a pseudocomplement; in fact, a complement of an element neednot be unique.

In a pseudocomplemented lattice L the set S(L) = {x∗ | x ∈ L} formsa lattice (called the skeleton of L), which is a ∧-subsemilattice of L and inwhich the join is defined by x � y = (x ∨ y)∗∗ = (x∗ ∧ y∗)∗. The lattice(S(L),∧,�) is in fact a boolean lattice [7, Theorem I.6.4]; this result wasfirst proved for complete distributive lattices by V. Glivenko [6].

A lattice L is called distributive if the identity

a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)

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holds for all a, b, c ∈ L. A pseudocomplemented distributive lattice L forwhich the skeleton S(L) is a sublattice of L is called a Stone lattice.

Let (G, ·, e) be a group (where e denotes the identity of G). Then the setL(G) consisting of all subgroups of G is a complete bounded lattice with topG and bottom {e}, called the subgroup lattice of G. The normal subgrouplattice N(G) of G is the modular sublattice of L(G) constituted by all normalsubgroups of G. Many classes of groups determined by different propertiesof their subgroup lattices or normal subgroup lattices (such as modularity,distributivity, complementation,. . . and so on) have been identified. In thepresent paper our aim is to investigate the groups G for which the latticeL(G) or N(G) is pseudocomplemented.

The paper is organized as follows. In Section 2 we study the class of finitePK-groups, i.e. the groups with pseudocomplemented subgroup lattice. InSection 3 we look at finite PKN-groups, i.e. the groups with pseudocomple-mented normal subgroup lattice. Section 4 deals with groups for which allsubgroups have pseudocomplemented lattices of normal subgroups. In the fi-nal section 5, finite groups whose normal subgroup lattices are Stone latticeswill be investigated.

Most of our notation is standard and will not be repeated here. Basicdefinitions and results on lattices and groups can be found in [3, 7] and [1, 18],respectively. For subgroup lattice concepts we refer the reader to [12, 17, 21].

We only mention that by a generalized quaternion group, we mean a groupof order 2t for some natural number t, defined by the presentation

Q2t = 〈a, b | a2t−2

= b2, a2t−1

= 1, b−1ab = a−1〉and not the more general notion of groups of order 4n of the form Q4n =〈a, b | an = b2, a2n = 1, b−1ab = a−1〉 which is used by some authors.

2 PK-groups

Definition 2.1. A group G is called a PK-group if its lattice of subgroupsL(G) is pseudocomplemented, and it is called a PKN-group if its lattice ofnormal subgroups N(G) is pseudocomplemented.

PK-groups were introduced and studied in [19] and [20]. Our goal in thissection is to give a complete classification of finite PK-groups.

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Theorem 2.2. Let G be a finite group of order 2tm with m odd. Then G isa PK-group if and only if it is cyclic or it is isomorphic to the direct productof a cyclic group of order m and a generalized quaternion group of order 2t.

We start with the following easy but important lemma.

Lemma 2.3. (i) Every subgroup of a PK-group is again a PK-group;

(ii) every cyclic group is a PK-group;

(iii) a group of order p2, p prime, is a PK-group if and only if it is cyclic;

(iv) every generalized quaternion group is a PK-group.

Proof. (i) Let G be PK-group and let H ≤ G. Let A be an arbitrarysubgroup of H and denote its pseudocomplement in G by A∗. ThenA∗ ∩ H is a pseudocomplement of A in H .

(ii) Let G be a finite cyclic group, and let A be an arbitrary subgroup ofG. Let s be the largest divisor of |G| coprime to |A|, and let A∗ be theunique subgroup of G of order s. Then A∗ is the pseudocomplement ofA in G.

Now let G be the infinite cyclic group (Z, +); then any two non-trivialsubgroups of G intersect non-trivially. So let A be an arbitrary subgroupof G, then A∗ = {0} is a pseudocomplement of A in G.

(iii) Let G be a PK-group of order p2, and assume that G is not cyclic.Then G ∼= Cp × Cp, hence G has p + 1 subgroups of order p. Take anyone of them, say A, and let A∗ be its pseudocomplement. Now taketwo distinct subgroups B1, B2 ≤ G of order p different form A; thenA ∩ Bi = 1, hence by the definition of A∗ we have Bi ≤ A∗ implyingG = 〈B1, B2〉 ≤ A∗ and hence A ∩ A∗ = A = 1, a contradiction.

(iv) Let G be a generalized quaternion group. It is well known that such agroup contains a unique involution z; see, for example, [1, Chapter 8,Exercise 3(7)]. But then every non-trivial subgroup A of G contains z,and hence A∗ = 1 is a pseudocomplement for A in G. �

Corollary 2.4. Let G be a finite PK-group. Then every Sylow subgroup ofG is either cyclic or generalized quaternion.

Proof. Let p be any prime divisor of |G|, and let S be a Sylow p-subgroupof G; by Lemma 2.3(i and iii), every subgroup of S of order p2 is cyclic.

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Equivalently, S has a unique subgroup of order p, since the center Z(S) ofS always contains a subgroup of order p. If p is odd, this implies that Sis cyclic, and if p = 2, this implies that S is either cyclic or generalizedquaternion; see, for example, [1, Chapter 8, Exercise 4]. �

Before we proceed, we show that the PK and PKN properties behave wellwith respect to direct products of groups of coprime order.

Proposition 2.5. Let G ∼= G1 × · · · × Gn with gcd(|Gi|, |Gj|) = 1 for alli, j ∈ {1, . . . , n} with i = j.

(i) G is a PK-group if and only if each Gi is a PK-group;

(ii) G is a PKN-group if and only if each Gi is a PKN-group.

Proof. We only prove (ii); the proof of (i) is similar. So assume first that G isa PKN-group, let i ∈ {1, 2, . . . , n} and let N �Gi. Then N �G, hence N hasa pseudocomplement N∗ in G; it is clear that N∗∩Gi is a pseudocomplementof N in Gi.

Conversely, assume that each Gi is a PKN-group, and let N � G; sincegcd(|Gi|, |Gj|) = 1 for all i, j ∈ {1, . . . , n} with i = j, we can write N =N1 × · · · × Nn, where Ni � Gi, for all i ∈ {1, . . . , n}. For each i, let N∗

i be apseudocomplement of Ni in Gi. Then N∗ = N∗

1 × · · · × N∗n is a pseudocom-

plement of N in G. �

Remark 2.6. It is clear from Lemma 2.3(iii) that the coprimeness conditioncannot be omitted.

To deal with non-solvable groups, we will make use of the following strongresult by M. Suzuki.

Theorem 2.7 (Suzuki [16]). Let G be a non-solvable finite group such thatevery Sylow subgroup of G is either cyclic or generalized quaternion. Then Gcontains a normal subgroup G1 such that [G : G1] ≤ 2 and G1

∼= Z×L, whereZ is a solvable group whose Sylow subgroups are all cyclic, and L ∼= SL(2, p)for some odd prime p.

We are now ready to prove Theorem 2.2.

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Proof of Theorem 2.2. Assume first that G is either cyclic or isomorphic tothe direct product of a cyclic group of order m and a generalized quaterniongroup of order 2t. Then it is clear from Lemma 2.3(ii and iv) together withProposition 2.5(i) that G is a PK-group.

Assume now that G is a PK-group, and suppose first that G is not solv-able. By Suzuki’s Theorem 2.7, there is an odd prime p such that G containsa subgroup L ∼= SL(2, p). By Lemma 2.3(i), L is a PK-group. We will showthat this leads to a contradiction. Indeed, let Z = Z(L) be the center of L;then Z = {1,−1}. Assume that Z∗ is the pseudocomplement of Z in L. Nowlet a be an arbitrary element of L of order p. Then 〈a〉∩Z = 1, hence a ∈ Z∗.Since L is generated by its elements of order p — in fact L = 〈( 1 1

0 1 ) , ( 1 01 1 )〉

— we obtain L ≤ Z∗ and hence Z ∩ Z∗ = Z = 1, a contradiction.

Hence G is solvable. By Corollary 2.4, every Sylow subgroup of G is eithercyclic or generalized quaternion. It is now sufficient to verify that every Sylowsubgroup of G is normal, since it then follows that G is the direct productof its Sylow subgroups, from which the statement in the theorem follows.

So suppose that G has a Sylow p-subgroup S which is not normal, andlet SG be the core of S in G. Since SG is a normal subgroup of G, we haveSG = S.

Case 1. SG = 1.

Let S∗G be a pseudocomplement of SG. We claim that

gcd(|SG|, |S∗G|) = 1 . (2.1)

Indeed, if p | |S∗G|, then there exists an M ≤ S∗

G with |M | = p. Let x ∈ Gbe such that M ≤ Sx. Thus M is the unique subgroup of order p in Sx.Hence M ≤ SG and so M ≤ SG ∩S∗

G = 1, a contradiction. We conclude thatequation (2.1) holds.

Since G is solvable, it follows from (2.1) that G has a Hall p′-subgroupH such that S∗

G ≤ H . On the other hand, SG ∩ H = 1, and hence H ≤ S∗G;

so S∗G = H . Now let a ∈ G. Since SG is a p-group and Ha is a p′-group,

we have SG ∩ Ha = 1; this implies Ha ≤ S∗G = H for all a ∈ G, and hence

H = S∗G is a normal Hall p′-subgroup of G.

Let H∗ be a pseudocomplement of H . Since H ∩ H∗ = 1 and H is theunique Hall p′-subgroup, H∗ must be a p-subgroup of G. So there is some

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x ∈ G such that H∗ ≤ Sx. Because H∩Sx = 1, we also have Sx ≤ H∗; henceSx = H∗. However, for every a ∈ G, we have H ∩ (H∗)a = 1 as well, hence(H∗)a ≤ H∗. We conclude that H∗ is normal and therefore S is normal, acontradiction.

Case 2. SG = 1.

We have 1 = SG =⋂

x∈G Sx =⋂

x∈G(S ∩ Sx). Since S has a uniquesubgroup of order p, this implies that there is an x ∈ G with S ∩ Sx = 1.Hence Sx ≤ S∗, where S∗ is a pseudocomplement of S. Now let H bean arbitrary Hall p′-subgroup of G; then S ∩ H = 1, therefore H ≤ S∗.We conclude that G = 〈Sx, H〉 ≤ S∗, hence S∗ = G. This contradictsS ∩ S∗ = 1. �

3 PKN-groups

In contrast to the situation of PK-groups, a (normal) subgroup of a PKN-group is not necessarily a PKN-group. For example, the dihedral group D8

of order 8 is a PKN-group, but has a normal subgroup C2 ×C2, which is nota PKN-group. This simple fact makes the study of PKN-groups considerablyharder.

We start with an easy but useful observation.

Proposition 3.1. Let G be a finite PKN-group. Then the center Z(G) of Gis cyclic.

Proof. Let H ≤ Z(G) be arbitrary; then H � G, so it has a pseudocom-plement H∗ in N(G). It is clear that Z(G) ∩ H∗ is a pseudocomplementof H in L(Z(G)). Hence Z(G) is a PK-group; the result now follows fromTheorem 2.2 since Z(G) is abelian. �

Proposition 2.5(ii) shows that the PKN property behaves well with re-spect to direct products of coprime order. For arbitrary direct products, westill have the following facts.

Lemma 3.2. Let G = G1 × · · · × Gn be a PKN-group. Then

(i) each Gi is a PKN-group;

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(ii) if N � G is such that N ∩ Gi = 1 for all i, then N = 1.

Proof. (i) Let i ∈ {1, 2, . . . , n} and let N � Gi. Then N � G, hence N hasa pseudocomplement N∗ in G; it is clear that N∗ ∩Gi is a pseudocom-plement of N in Gi.

(ii) Let N∗ be the pseudocomplement of N in G. Since N ∩ Gi = 1, wehave Gi ≤ N∗ for all i, hence N∗ = G and since N∩N∗ = 1 this impliesN = 1. �

Remark 3.3. By a well-known result of J. Wiegold [24], a group has acomplemented lattice of normal subgroups (such a group is called an nD-group – see [22]) if and only if it is a direct product of simple groups. So annD-group is not always a PKN-group. A sufficient condition for a direct pro-duct G of simple groups to become a PKN-group is that no two its minimalabelian factors are isomorphic. In this case N(G) is a boolean lattice (see[23, Lemma 1]) and hence a pseudocomplemented lattice.

In any normal subgroup lattice N(G) we have 1∗ = G and G∗ = 1.A special type of PKN-groups G is obtained when the skeleton S(N(G))only contains the subgroups 1 and G.

Definition 3.4. We say that a PKN-group G is elementary if S(N(G)) ={1, G}.

A characterization of finite elementary PKN-groups is given by the fol-lowing result.

Proposition 3.5. A finite group G is an elementary PKN-group if and onlyif it is a monolithic group, i.e. a group with a unique minimal normal sub-group.

Proof. Suppose that G is an elementary PKN-group and assume that it hasat least two minimal normal subgroups M1 and M2. Because M1 ∩ M2 = 1,it follows that M2 ≤ M∗

1 and therefore M∗1 = 1, a contradiction. Conversely,

suppose that G is a monolithic group and let M be its unique minimal normalsubgroup. Then for every H � G with H = 1, we have M ≤ H . This showsthat H∗ = 1. Hence G is an elementary PKN-group. �

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Remark 3.6. This result is no longer true for infinite groups, since an infinitegroup may have no minimal normal subgroups at all. For example, theinfinite cyclic group (Z, +) is elementary PKN, but has no minimal normalsubgroups.

Remark 3.7. Using Proposition 3.5, many classes of elementary PKN-groups can be obtained. For example, any group whose normal subgrouplattice is a chain is an elementary PKN-group. In particular, simple groups,symmetric groups, cyclic p-groups or finite groups of order pnqm (p, q distinctprimes) with cyclic Sylow subgroups and trivial center [12, Exercise 3, p. 497]are all elementary PKN-groups; see also Theorem 5.9 and Theorem 5.12 be-low. We also mention that any proper semidirect product of type G = 〈a〉M ,where M is a maximal normal subgroup of G having a fully ordered latticeof normal subgroups, is an elementary PKN-group.

Obviously, elementary PKN-groups G have the property that the latticeS(N(G)) is a sublattice of N(G). Our next result gives a characterization offinite PKN-groups for which this property holds.

Proposition 3.8. For a finite PKN-group G, the following two propertiesare equivalent:

(a) S(N(G)) is a sublattice of N(G);

(b) G is a direct product of elementary PKN-groups.

Proof. For each normal subgroup N � G, we will denote its pseudocomple-ment in G by N∗. We claim that (a) is equivalent with the condition

(A ∩ B)∗ = A∗B∗ for all A, B ∈ S(N(G)) . (3.1)

Indeed, observe that S(N(G)) = {A � G | A∗∗ = A}. If S(N(G)) is asublattice of N(G), then for all A, B ∈ S(N(G)), we have AB = A � B =(A∗ ∩ B∗)∗, so by replacing A with A∗ and B with B∗ we obtain (3.1).Conversely, if (3.1) holds, then for all A, B ∈ S(N(G)), we have AB =A∗∗B∗∗ = (A∗ ∩ B∗)∗ ∈ S(N(G)), so S(N(G)) is a sublattice of N(G).

(a)⇒(b). To prove (b), we will use induction on |G|. Let M1, M2, . . . , Ms

be the minimal normal subgroups of G. If s = 1, then G itself is an

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elementary PKN-group, so we may assume s ≥ 2. Then M∗1 ∩M∗∗

1 = 1,and by equation (3.1),

G = 1∗ = (M∗1 ∩ M∗∗

1 )∗ = M∗∗1 M∗∗∗

1 = M∗∗1 M∗

1 ,

hence G is the direct product of M∗∗1 and M∗

1 . In particular, everyminimal normal subgroup of M∗∗

1 is also a minimal normal subgroupof G.

Now Mi ≤ M∗1 , for all i ∈ {2, . . . , s}, hence Mi ∩ M∗∗

1 = 1. ThereforeM∗∗

1 has a unique minimal normal subgroup, namely M1, and henceM∗∗

1 is an elementary PKN-group.

By Lemma 3.2(i), M∗1 is again a PKN-group. It remains to show that

S(N(M∗1 )) is a sublattice of N(M∗

1 ), since the result will then followby the induction hypothesis. We have

S(N(M∗1 )) = {K∗ ∩ M∗

1 | K ∈ N(G)} = {A ∩ M∗1 | A ∈ S(N(G))} .

For any two elements A ∩ M∗1 and B ∩ M∗

1 of S(N(M∗1 )), we have

(A ∩ M∗1 ) ∩ (B ∩ M∗

1 ) = (A ∩ B) ∩ M∗1 ∈ S(N(M∗

1 )) ,

(A ∩ M∗1 )(B ∩ M∗

1 ) = (AB) ∩ M∗1 ∈ S(N(M∗

1 )) ,

where we have used the fact that S(N(G)) is a distributive lattice. Thisshows that S(N(M∗

1 )) is a sublattice of N(M∗1 ), as claimed.

(b)⇒(a). Let G = G1 × · · · × Gn, where each Gi is an elementary PKN-group. Let K ∈ S(N(G)) be arbitrary, and let i ∈ {1, . . . , n}. SinceK ∩ Gi ∈ S(N(Gi)) = {1, Gi}, we have either K ∩ Gi = 1 (and thenGi ≤ K∗) or Gi ≤ K (and then K∗ ∩ Gi = 1). Let

X = {i ∈ {1, . . . , n} | K ∩ Gi = 1} = {i ∈ {1, . . . , n} | Gi ≤ K∗} ,

Y = {i ∈ {1, . . . , n} | K∗ ∩ Gi = 1} = {i ∈ {1, . . . , n} | Gi ≤ K} ,

and write A =∏

i∈X Gi and B =∏

i∈Y Gi. Then G = A × B; byLemma 3.2(i), both A and B are PKN-groups. Observe that B ≤ K.By the modular law,

K = K ∩ G = K ∩ (AB) = B(K ∩ A) .

However, by Lemma 3.2(ii) applied on the PKN-group A, we haveK ∩ A = 1. Hence K = B =

∏i∈Y Gi.

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Now take K1, K2 ∈ S(N(G)); then there are subsets Y and Z of{1, . . . , n} such that K1 =

∏i∈Y Gi and K2 =

∏i∈Z Gi. Since S(N(G))

is always a ∧-subsemilattice of N(G), we have K1 ∩ K2 ∈ S(N(G))(in fact K1 ∩ K2 =

∏i∈Y ∩Z Gi). On the other hand, it is obvious that

K1K2 =∏

i∈Y ∪Z Gi ∈ S(N(G)). Hence S(N(G)) is a sublattice ofN(G). �

Remark 3.9. If the equivalent properties of Proposition 3.8 are satisfied,then S(N(G)) is in fact a direct product of chains of length 1; this followsfrom the proof of (b)⇒(a).

A complete classification of finite PKN-groups seems to be out of reachat this point, but we are able to classify all finite nilpotent PKN-groups.

Theorem 3.10. Let G be a finite nilpotent group. Then G is a PKN-groupif and only if its center Z(G) is cyclic.

Proof. If G is a PKN-group, then Z(G) is cyclic by Proposition 3.1. Soassume that G is a finite nilpotent group with cyclic center. Then G isisomorphic to the direct product of its Sylow subgroups, G ∼= ∏k

i=1 Gi, whereeach Gi is a pi-group with cyclic center. By Proposition 2.5(ii), G is a PKN-group if and only if each Gi is a PKN-group.

So we may assume without loss of generality that G is a p-group withcyclic center Z(G). But then the unique subgroup A of Z(G) of order p isin fact the unique minimal normal subgroup of G. By Proposition 3.5, G isa PKN-group (in fact an elementary PKN-group). �Corollary 3.11. Let G be a finite nilpotent PKN-group. Then

(i) G is the direct product of elementary PKN-groups;

(ii) S(N(G)) is a sublattice of N(G).

Proof. Part (i) follows from the fact that a PKN-p-group is elementary PKN;part (ii) then follows from Proposition 3.8. �

4 PKN∗-groups

As we have observed, the class of PKN-groups is not closed under (normal)subgroups. It therefore makes sense to introduce the following class of groups.

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Definition 4.1. A group G is called a PKN∗-group if all of its subgroups arePKN-groups.

Our goal in this section is to study finite PKN∗-groups; in particular, wewill classify finite solvable PKN∗-groups.

Theorem 4.2. Let G be a finite solvable group. Then G is a PKN∗-groupif and only if all Sylow subgroups of G of odd order are cyclic and all Sylow2-subgroups of G are cyclic or generalized quaternion.

Proof. Assume first that G is a PKN∗-group; then every subgroup of orderp2 (with p prime) is itself a PKN-group, and hence such a subgroup is cyclicby Lemma 2.3(iii). As in the proof of Corollary 2.4, it follows that everySylow subgroup of G is either cyclic or generalized quaternion.

Conversely, suppose that G is a solvable group such that all Sylow sub-groups are either cyclic or generalized quaternion. Let |G| = pα1

1 pα22 · · ·pαk

k bethe decomposition of |G| as product of prime factors. For each i ∈ {1, . . . , k},let Hi be a Sylow pi-subgroup of G. We claim that for every two normal sub-groups U, V � G, the equivalence

U ∩ V = 1 ⇐⇒ gcd(|U |, |V |) = 1 (4.1)

holds. Indeed, write |U | =∏k

i=1 pβii and |V | =

∏ki=1 pγi

i , and denote theSylow pi-subgroups of U and V by Ui and Vi, respectively. Of course, ifgcd(|U |, |V |) = 1, then U ∩ V = 1. Assume gcd(|U |, |V |) = 1, then thereis an index i ∈ {1, . . . , k} such that βi ≥ 1 and γi ≥ 1. Then there existx, y ∈ G such that Ux

i ≤ Hi and V yi ≤ Hi. Thus Ux

i ∩ V yi ≤ Hi. Since

Hi is either cyclic or generalized quaternion, it has exactly one subgroup oforder pi. This implies that Ux

i ∩ V yi = 1 and so U ∩ V = 1. This proves the

equivalence (4.1).

Now, suppose without loss of generality that βi = 0 for all i ∈ {1, . . . , s}and βs+1 = βs+2 = · · · = βk = 0. Since G is solvable, it contains a sub-group W of order

∏ki=s+1 pαi

i ; let U∗ = coreG(W ). Then by (4.1), U∗ is apseudocomplement of U in G. �

Remark 4.3. The ZM-groups, i.e. the groups with all Sylow subgroupscyclic, are always solvable, and hence it follows from Theorem 4.2 that allZM-groups are PKN∗-groups. Another important (lattice theoretical) prop-erty of the ZM-groups is that these groups are exactly the finite groups whose

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poset of conjugacy classes of subgroups forms a distributive lattice [4, The-orem A].

Not all PKN∗-groups are solvable. We will show that the group SL(2, p)with p prime is always a PKN∗-group. We start with a lemma.

Lemma 4.4. Let G = SL(2, p) for some prime p, and let H be a non-solvablesubgroup of G. Then H ∼= SL(2, 5).

Proof. Let Z be the center of G, then |Z| = 2 and G/Z ∼= PSL(2, p). FromDickson’s list of subgroups of PSL(2, q) (see, for example, [8, II, Haupt-satz 8.27]), we deduce that the only possible non-solvable subgroup of PSL(2, p)is Alt(5), hence HZ/Z ∼= Alt(5). In particular, |H| ∈ {60, 120}. Recall thatSL(2, p) contains a unique involution, so |H| = 60 would imply Z ≤ H and|HZ/Z| = 30; hence |H| = 120. However, of the three non-solvable groupsof order 120 —SL(2, 5), Sym(5) and C2 × Alt(5)— only SL(2, 5) contains aunique involution. �

Theorem 4.5. Let G = SL(2, p) for some prime p. Then G is a PKN∗-group.

Proof. For p ∈ {2, 3}, it is easy to check directly that G is a PKN∗-group; sowe may assume p ≥ 5. Then G contains a unique non-trivial normal propersubgroup (namely its center), and hence it is an (elementary) PKN-group.

Now let H < G be an arbitrary proper subgroup. It is well known thatthe Sylow subgroups of SL(2, p) are either cyclic or generalized quaternion—see, for example, [8, II, Satz 8.10(a)]— and hence the same is true for H .If H is solvable, then it follows from Theorem 4.2 that H is a PKN-group.If H is non-solvable, then Lemma 4.4 implies H ∼= SL(2, 5), and by the firstparagraph of the proof, H is a PKN-group. �

5 Groups whose normal subgroup lattices are

Stone lattices

In this final section we present some results concerning finite groups G forwhich the lattice N(G) is a Stone lattice, i.e. a distributive pseudocomple-mented lattice such that S(N(G)) is a sublattice of N(G).

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We mention that the structure of groups with distributive lattices of nor-mal subgroups, the so-called DLN-groups, is not known, but there are somecharacterizations of these groups (see [10] or [12, §9.1]). We only mentionthe following theorem, which is a generalization of [10, Theorem 4.1].

Theorem 5.1. Let G = G1 × · · · × Gk be a finite group. Then G is aDLN-group if and only if each direct factor Gi is a DLN-group and for alli = j, there are no central chief factors in Gi and in Gj the orders of whichcoincide.

We are now able to establish the following characterization of finite groupswhose normal subgroup lattice is a Stone lattice.

Theorem 5.2. Let G be a finite group. Then N(G) is a Stone lattice if andonly if G is a direct product G = G1 × · · · × Gk of monolithic DLN-groupsGi, and for all i = j, there are no central chief factors in Gi and in Gj theorders of which coincide.

Proof. Assume first that N(G) is a Stone lattice. Then G is a PKN-groupand S(N(G)) is a sublattice of N(G), so by Proposition 3.8, G is a directproduct of monolithic groups: G = G1 × · · · × Gk. Since G is a DLN-group,the conclusion now follows from Theorem 5.1.

Conversely, assume that G = G1×· · ·×Gk, where each Gi is a monolithicDLN-group, and for all i = j, there are no central chief factors in Gi andin Gj the orders of which coincide. Then by Theorem 5.1, N(G) is a dis-tributive lattice. We now claim that G is a PKN-group. Indeed, let N � G.By distributivity of N(G), we have N =

∏ki=1(N ∩ Gi). Without loss of

generality, we may assume that N ∩Gi = 1 for i ∈ {1, . . . , s} and N ∩Gi = 1for i ∈ {s + 1, . . . , k}; in particular, N ≤ ∏s

i=1 Gi. Let N∗ =∏k

i=s+1 Gi;we claim that N∗ is a pseudocomplement of N in N(G). Indeed, we clearlyhave N ∩ N∗ = 1. Assume that A � G is such that N ∩ A = 1. For eachi ∈ {1, . . . , s}, the group Gi is monolithic, so since N ∩Gi = 1 we must haveA ∩ Gi = 1. But then by distributivity, A =

∏ki=1(A ∩ Gi) ≤ N∗, which

proves that N∗ is a pseudocomplement of N as claimed. It now follows fromProposition 3.8 that S(N(G)) is a sublattice of N(G), and hence N(G) is aStone lattice. �

Remark 5.3. The structure of finite monolithic DLN-groups is not knowneither. This class of groups obviously includes the cyclic p-groups, but there

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exist a lot of non-cyclic groups which are monolithic DLN-groups (for exam-ple, the non-cyclic elementary PKN-groups presented in Remark 3.7). Nev-ertheless, it is easy to see that the cyclic p-groups are the only finite p-groupsthat are monolithic DLN-groups. Moreover, because a direct product of kcyclic p-groups is a PKN-group if and only if k = 1, the cyclic p-groups arealso the only finite p-groups G for which N(G) is a Stone lattice.

The previous remark allows us to classify finite nilpotent groups whosenormal subgroup lattice is a Stone lattice.

Corollary 5.4. Let G be a finite nilpotent group. Then N(G) is a Stonelattice if and only if G is cyclic.

Proof. Since G is nilpotent, it can be written as a direct product of its Sylowsubgroups: G = G1 × · · · × Gk. If G is cyclic, then each of the groupsGi is a cyclic pi-group, and hence a monolithic DLN-group; it follows fromTheorem 5.2 that N(G) is a Stone lattice.

Conversely, assume that N(G) is a Stone lattice. Since the orders of theSylow subgroups are coprime, Theorem 5.2 implies that each of the latticesN(Gi) is a Stone lattice; by Remark 5.3, each Gi is cyclic. Hence G is alsocyclic. �

We will now investigate finite groups for which the normal subgroup lat-tice of every subgroup is a Stone lattice. We start with a reduction; let usfirst recall the following definition which we have encountered in Remark 4.3.

Definition 5.5. A finite group G is called a Zassenhaus metacyclic group,or ZM-group for short, if all Sylow subgroups of G are cyclic.

Proposition 5.6. Let G be a finite group. Then the following two propertiesare equivalent:

(a) N(H) is a Stone lattice for all subgroups H of G;

(b) G ∼= G1 × · · · × Gk, where the Gi are monolithic ZM-groups of coprimeorders, with the property that N(Hi) is a Stone lattice for all subgroupsHi of Gi.

Proof. (a)⇒(b). Let H be a Sylow subgroup of G. Then N(H) is a Stonelattice and therefore, by Corollary 5.4, H is cyclic. Thus G is a ZM-group. Since N(G) is a Stone lattice, Theorem 5.2 implies that G

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can be written as a direct product of monolithic DLN-groups Gi, i ∈{1, . . . , k}. Since every subgroup of a ZM-group is itself a ZM-group,the Gi are in fact ZM-groups. Suppose that there are i = j suchthat |Gi| and |Gj | have a common prime divisor p. Then there are asubgroup Mi or order p in Gi, and a subgroup Mj of order p in Gj . Byour hypothesis, the normal subgroup lattice of H = Mi×Mj

∼= Cp×Cp

is a Stone lattice, a contradiction.

(b)⇒(a). Let H be an arbitrary subgroup of G. Since the direct factors Gi

have coprime orders, H ∼= H1×· · ·×Hk for certain subgroups Hi ≤ Gi.Clearly H is a ZM-group, so by Remark 4.3 it is a DLN-group. SinceN(Hi) is a Stone lattice for each i, Proposition 3.8 implies that each Hi

is a direct product of elementary PKN-groups, and hence the same istrue for H . Moreover, each Hi is a PKN-group, and since the Hi havecoprime orders, Proposition 2.5(ii) implies that H is a PKN-group. Weconclude that N(H) is a Stone lattice. �

In view of Proposition 5.6(b), we will now digress on monolithic ZM-groups. The structure of ZM-groups has been completely determined byZassenhaus.

Definition 5.7. A triple (m, n, r) satisfying the conditions

gcd(m, n) = gcd(m, r − 1) = 1 and rn ≡ 1 (mod m)

will be called a ZM-triple, and the corresponding group

⟨a, b | am = bn = 1, b−1ab = ar

⟩

will be denoted by ZM(m, n, r).

Theorem 5.8 (Zassenhaus). Let G be a ZM-group. Then there exists a ZM-triple (m, n, r) such that G ∼= ZM(m, n, r). We have |G| = mn and G′ = 〈a〉(so |G′| = m), and G/G′ is cyclic of order n.

Conversely, every group isomorphic to ZM(m, n, r) is a ZM-group.

Proof. See, for example, [8, IV, Satz 2.11]. �

We can now determine the structure of monolithic ZM-groups.

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Theorem 5.9. Let G ∼= ZM(m, n, r) be a ZM-group. Then the followingproperties are equivalent:

(a) G is monolithic;

(b) either m = 1 and n is a prime power, or m is a prime power and rd ≡ 1(mod m) for all 1 ≤ d < n;

(c) either |G| is a prime power, or |G′| is a prime power and Z(G) = 1.

Proof. (a)⇒(b). Assume that G ∼= ZM(m, n, r) is monolithic. If m = 1, thenG is cyclic of order n; for each prime divisor p of n, G has a minimalnormal subgroup of order p; hence n is a prime power.

So assume that m = 1, and let p be a prime divisor of m. By thedefining relations of G in Definition 5.7, we see that every subgroupof 〈a〉 is normalized by b and thus is a normal subgroup of G. Inparticular, G has a minimal normal subgroup of order p.

Hence m can have only one prime divisor, i.e. m is a prime power; let Nbe the minimal normal subgroup contained in 〈a〉. Suppose now thatthere is some d < n such that rd ≡ 1 (mod m). Since rn ≡ 1 (mod m)as well, we may in fact assume that d | n; assume furthermore that n/dis prime. Let H = 〈bd〉; this is a subgroup of G of prime order. By thedefining relation ab = bar, we get

abd = bdard

= bda

since rd ≡ 1 (mod m); this shows that bd ∈ Z(G), so in particular His a minimal normal subgroup of G, different from N . This contradictsthe fact that G is monolithic.

(b)⇒(c). If m = 1 and n is a prime power, then |G| = mn is a prime power.So assume that m = |G′| is a prime power and rd ≡ 1 (mod m) for all1 ≤ d < n. Let g = bsat with 0 ≤ s ≤ n − 1 and 0 ≤ t ≤ m − 1 be anarbitrary element of G; we compute

[g, a] = a1−rs

, [g, b] = at(r−1) .

Assume now that g ∈ Z(G); then we must have rs ≡ 1 (mod m) andm | t(r−1). Our assumption rd ≡ 1 (mod m) for all 1 ≤ d < n impliess = 0; the fact that gcd(m, r − 1) and t < m implies t = 0. HenceZ(G) = 1.

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(c)⇒(a). If |G| is a prime power, then G is a cyclic group of prime powerorder and hence it is monolithic. So assume that |G′| is a prime powerand Z(G) = 1. Then G′ has a unique subgroup P of prime order,which is therefore a minimal normal subgroup of G. Let N � G bean arbitrary non-trivial normal subgroup of G; it is sufficient to showthat N ∩ G′ = 1, since this will imply that P ≤ N . So suppose thatN∩G′ = 1. Then [N, G] = 1, and hence N ≤ Z(G). But now Z(G) = 1implies N = 1, a contradiction. �

We now claim that every monolithic ZM-group has the property that thenormal subgroup lattice of any subgroup is a Stone lattice. This statementis obvious when the group is cyclic of prime power order, so we only considernon-abelian ZM-groups.

Theorem 5.10. Let G ∼= ZM(m, n, r) be a monolithic non-abelian ZM-groupwith m = pk for some prime p and some k ≥ 1. Then

(i) n | p − 1;

(ii) the order of each element of G is a divisor of m or a divisor of n;

(iii) N(H) is a Stone lattice for each subgroup H of G. More precisely, everysubgroup of G is either cyclic or monolithic.

Proof. Let G ∼= ZM(m, n, r) be a monolithic ZM-group with generators aand b as in Definition 5.7. By Theorem 5.9, m = pk for some prime p andsome number k ≥ 1 since we assume G to be non-abelian.

(i) By Theorem 5.9(b), the order of r modulo pk is precisely n. In particu-lar, n | φ(pk) = pk−1(p− 1), where φ is the Euler totient function; sincegcd(n, p) = 1, we have n | p − 1.

(ii) By the defining relations in Definition 5.7, it is not very hard to com-pute, using induction on d, that

(bsat)d = bsd at(1+rs+r2s+···+r(d−1)s) , (5.1)

for all natural numbers s, t, d. Suppose that G contains an elementg = bsat of order pq, where q is a prime dividing n. Then bs must haveorder q, hence we have s = c · n/q for some c ∈ {1, . . . , q − 1}. Let

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x = rs; then the order of x modulo pk is q. We have

gq = at(1+x+x2+···+x(q−1)) , gpq = at(1+x+x2+···+x(pq−1)) .

Since g has order pq, this implies

t(1 + x + x2 + · · ·+ x(q−1)

) ≡ 0 (mod pk) , (5.2)

t(1 + x + x2 + · · ·+ x(pq−1)

) ≡ 0 (mod pk) . (5.3)

Also observe that by (i), we have q | p − 1, so in particular xp ≡ x(mod pk). Let y = 1 + x + · · · + xp−1; then y(x − 1) = xp − 1 ≡ x − 1(mod pk), hence pk | (y − 1)(x − 1). However, x ≡ 1 (mod pk), andtherefore p | y − 1. In particular, gcd(y, p) = 1.

Since xp ≡ x (mod pk), we have

1 + x + x2 + · · ·+ x(pq−1)

=(1 + x + x2 + · · · + x(p−1)

) (1 + xp + x2p + · · ·+ x(q−1)p

)

≡ y(1 + x + x2 + · · · + x(q−1)

)(mod pk) ;

since gcd(y, p) = 1, this implies the equivalence

t(1 + x + x2 + · · · + x(pq−1)

) ≡ 0 (mod pk)

⇐⇒ t(1 + x + x2 + · · · + x(q−1)

) ≡ 0 (mod pk) .

This contradicts equations (5.2) and (5.3), and (ii) follows.

(iii) Let H be an arbitrary subgroup of G. If H is abelian, then it is cyclic,so Corollary 5.4 implies that N(H) is a Stone lattice. So we may assumethat H is non-abelian; in particular, the derived subgroup H ′ is non-trivial, and since H ′ ≤ G′ = 〈a〉, this implies that g = apk−1 ∈ H , andtherefore p | |H|. If p were the only prime dividing |H|, then H wouldbe cyclic, contradicting the fact that H is non-abelian; hence there issome prime q | n with q | |H|. Let h ∈ H be an element of order q.

Suppose that Z(H) = 1. If Z(H) contains an element x of order p, thenxh has order pq, contradicting (ii). If Z(H) does not contain elementsof order p, then it must contain an element y of order q′ for someprime q′ | n, but then yg has order pq′, again contradicting (ii). Weconclude that Z(H) = 1, and hence by Theorem 5.9, H is a monolithicZM-group; in particular, N(H) is a Stone lattice. �

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We now obtain the complete classification of all groups with the propertythat each subgroup has a Stone normal subgroup lattice as an easy corollary.

Theorem 5.11. Let G be a finite group. Then the following two propertiesare equivalent:

(a) N(H) is a Stone lattice for all subgroups H of G;

(b) G ∼= G1 × · · · × Gk, where the Gi are monolithic ZM-groups of coprimeorders.

Proof. This follows immediately from Proposition 5.6 and Theorem 5.10. �

We end this section by classifying all finite groups such that every sub-group is monolithic.

Theorem 5.12. Let G be a finite group; then the following two propertiesare equivalent:

(a) every subgroup of G is monolithic;

(b) either G is a cyclic group of order pk, or G is a ZM-group of order pkq�

with Z(G) = 1, where p, q are distinct primes.

Proof. (a)⇒(b). Assume that every subgroup of G is monolithic. Then everySylow subgroup of G is cyclic, hence G is a monolithic ZM-group, sayG ∼= ZM(m, n, r). Assume that |G| is not a prime power; then byTheorem 5.9, Z(G) = 1 and m is a prime power pk. On the otherhand, the group 〈b〉 is a cyclic group of order n; such a group can onlybe monolithic if n is a prime power q�.

(b)⇒(a). If G is a cyclic group of prime power order, then each of its sub-groups is monolithic; so assume that G is a ZM-group of order pkq�

with Z(G) = 1. By Theorem 5.9, G is monolithic. Now let H be anarbitrary subgroup of G. Then by Theorem 5.10(iii), H is cyclic ormonolithic. If it is monolithic, we are done, so assume H is cyclic. ByTheorem 5.10(ii), the order of each element is either a power of p or apower of q, so it follows that H is a cyclic p-group or a cyclic q-group.In both cases, H is monolithic. �

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Tom De MedtsDepartment of Pure Mathematicsand Computer AlgebraGhent UniversityGalglaan 2, B-9000 Gent, Belgiume-mail: tdemedts@cage.UGent.be

Marius TarnauceanuFaculty of Mathematics“Al.I. Cuza” UniversityIasi, Romaniae-mail: tarnauc@uaic.ro

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